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Energy conservation problem

  1. Apr 16, 2013 #1
    1. The problem statement, all variables and given/known data

    see link for a picture of the situation, http://i50.tinypic.com/n5g3z5.jpg

    2. Relevant equations

    (1/2)mv^(2) , (1/2)k*l^(2)

    3. The attempt at a solution

    Total energy of the system at the given point of time is

    mv^(2)/2+kl^(2)/2 = 0.6*100/2+100*0.25/2 = 30+12.5 = 42.5 J

    At the maximum and minimum values of l, v = 0

    => kl2/2 = 42.5

    => l = √(85/k)

    => l = ±0.922 m

    maximum l = 0.922m and minimum l = -0.922m

    Not terribly confident in this approach however, any guidance would be appreciated. Thanks!
     
  2. jcsd
  3. Apr 16, 2013 #2
    Is angular momentum conserved here?
     
  4. Apr 16, 2013 #3
    yes indeed, so this is a conservation of angular momentum problem.
     
    Last edited: Apr 16, 2013
  5. Apr 16, 2013 #4
    Is this enough to get on with the solution?
     
  6. Apr 16, 2013 #5
    I'm now reading up on angular momentum, but please if you have some tips to get the ball rolling that would be greatly appreciated.
     
  7. Apr 16, 2013 #6
    One thing you need to think about is the angle of the velocity vector with the string at the min/max positions. You need the angle to compute the angular momentum.
     
  8. Apr 17, 2013 #7
    Ok so far i have

    (1/2)k(xf^(2)-xi^(2)) = mv*l*sin(60) , I have taken xf=0.5m, so now i solve for xi.

    ((-(0.6)(10)sin(60)*2)/100) +0.5
    which gives me xi = 0.44m

    am i on the right track with this problem? any guidance would be appreciated, thanks!
     
  9. Apr 17, 2013 #8
    I do not understand what that equation means. The right hand side is the initial angular momentum. But what is on the left hand side? Is that the potential energy of the spring? How could you possibly equate energy and angular momentum?
     
  10. Apr 17, 2013 #9
    ok to start since angular momentum is conserved do i start out by writing the following equation?
    m*v1*l1*sin(60) = m*v2*l2sin(a)=m*v3*l3sin(b),
    l2 = min and l3 = max
     
  11. Apr 17, 2013 #10
    I'm guessing i need to use r-polar coordinates here also?
     
  12. Apr 17, 2013 #11
    Think about what I wrote in #6.
     
  13. Apr 17, 2013 #12
    ok so m*v1*l1*sin(60)-m*v3*l3sin(a) =0

    divide through by m*v3*l3,
    i get (mv1l1)/(mv3l3)(sin(60)-sin(a)) therefore 'a' must be equal to 60 at max L?
     
  14. Apr 17, 2013 #13
    I do not see why that would be true.

    Have you heard about radial and azimuthal components of velocity?
     
  15. Apr 17, 2013 #14
    no i haven't
     
  16. Apr 17, 2013 #15
    You mentioned polar coordinates earlier. Do you know what the polar-coordinates representation of a velocity vector looks like?
     
  17. Apr 17, 2013 #16
    yes indeed, you have a velocity in the theta direction and a velocity in the radial direction where the square root of the two terms squared are then the total magnitude of the velocity
     
  18. Apr 17, 2013 #17
    Very well. Now think about the radial velocity. When the radius is at a minimum or at a maximum, what can you say about the radial velocity?
     
  19. Apr 17, 2013 #18
    the radial velocity will be equal to 0, so v(theta) when l is at a maximum will be the velocity of the mass?
     
  20. Apr 17, 2013 #19
    oh no velocity in the radial direction for both L max/min will be 0
     
  21. Apr 17, 2013 #20
    And if the radial velocity at the max/min radii is zero, what is the overall direction of the velocity? What is its angle with the radius at min/max?
     
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